Triac - Power dissipated in light bulb

[Jerremy_S]

Hi All,

Having given a= π/3 Can you explain why b= 5π/3 ?

And Also, where this integral formula comes from? What is it called?

I am going back and forth through my textbooks and I cannot find anything.
A push in the right direction would be appreciated:)

Thanks

 

[gneill]

Hi All,

Having given a= π/3 Can you explain why b= 5π/3 ?

If you look at the actual conduction periods for the firing angle you'll see that conduction takes place in two separate periods from π/3 to π, and 4π/3 to 2π. Rather than write separate integrals for each, it is convenient to make use of the symmetry of the sine function whereby the area under the curve from π to 5π/3 is the same as the area from 4π/3 to 2π. Thus a single integral over the angles from π/3 to 5π/3 is equivalent to the two separate integrals.

And Also, where this integral formula comes from? What is it called?
View attachment 110276

That's the power over one period. It stems from the power relationship P = V²/R.